-Xi∕n∣<ξn}2n ʃj/n^/ (Xs Xj/n) σ(Xs)dW1 ,s
r- L(n-1) rj / yn-1 1 rl
ym ^ j = 1 {∣x3∕n
n ⅛ I
i =1 ∖
<-
Уn-1 1
2=3=1 1 {∣Xj∕n-xi∕n<M
S/
Gn,m,r
L ( n-1) rj yn-11
j = 1 {lxj∕n
-Xi∕n∖<ξn} 2 n ʃj/n ) ( Xs Xj/n ) μ ( Xs )d s
∑n-1 1
4.
j =1 1 {∣Xj∕n-Xi∕n∣<ξn}
SZ
Hn,m,r
L(n-1)rj / yn-11
+ ∖ m ^ j = 1 {\Xj/n
n ⅛ I
г=1
S-
-X,,n∖<ξn}n (ʃj/"11 /n (σ2(Xs) - σ2(Xi/n)) ds
∑n-1
j =1 1 {∣Xj∕n-Xi∕n∣<ξn}
--------S/----------------------------
Dn,m,r
Thus, we need to show that Gn,m,r, Hn,m,r and Dn,m,r are oP(1).
Now, because of Lemma 1,
Dn,m,r ≤ √m sup Iσ2(Xs) - σ2(Xτ)∣
∖Xs-Xτ ∣≤ξn
≤ √m sup IVσ2(Xτ)∣ sup ∖Xs — Xτ∣
τ∈ [0,1] ∖Xs-Xτ ∣≤ξn
= O (√m) Oa.s. (nε/2)Oa.s(ξn) = oa.s. (1),
(15)
(16)
provided that m1 /2nε/2ξn → 0. Since m = o(n1 ε), then
Oa.s ( √mnε/2ξn) = 0a.s (n1 /2ξn),
which approaches zero almost surely.
As for Gn,m,r, by the proof of Step 1 of Theorem 1, part(i)a, below, (√n∕√m)Gn,m,r = Gn,r
converges in distribution and so it’s Op(1); therefore Gn,m,r = oP(1), given that m∕n → 0, as
m,n → ∞.
Finally, given the continuity of μ(■ ),
H[snpι,r ∣ ≤ √m sup ∖μ(Xs) ∣ sup 1 Xs — Xi/n1
s∈ [0,1] |i/n-s∖≤ 1 /n
s∈[0,1]
= √mOasi( (nε/4)Oa.s. (n-1 /2 log n) = oa.s. (1). (17)
In fact, because of the modulus of continuity of a diffusion (see McKean, 1969, p.96),
sup 1 Xs — Xi/n 1 = Oa.s. n-~1 /2 logn) ,
[i/n-s^i/n
s∈[0,1]
and n1 /2-ε+ε/4n-1 /2 log n = n-3ε/4 log n → 0. Therefore, the statement follows. ■
We can now prove Theorem 1.
19